# The particle in a two and a three dimensional box: degeneracy of energy levels

## Course Contents

The particle in a two and a three dimensional box: Degeneracy of energy levels

Lecture Objectives

1. To introduce a model problem in quantum mechanics known as the particle in a two or three dimensional box with infinite-potential boundary.

2. To demonstrate the solution of the Schrödinger equation using the above model and using elementary differential equations with specific boundary conditions.

3. To highlight the additional consequence of this model leading to degeneracy of wave functions and the energy states.

4. To highlight quantization of energy as a result of boundary conditions applied to Schrödinger equation and probability interpretation for wave functions with multiple degrees of freedom by way of defining the appropriate 'area' and 'volume' elements along with the squares of the wave function.

5. To illustrate graphically the solutions for the wave functions and their squares using surface plots.

Lecture Outcomes

1. You will be able to solve a simple differential equation such as

${\color {Black} \left( \frac{\partial ^2}{ \partial x^2} + \frac{\partial ^2}{ \partial y^2} \right) \psi (x,y) + k^2 \psi (x,y) = 0 \,}$ or

${\color {Black} \left( \frac{\partial ^2}{ \partial x^2} + \frac{\partial ^2}{ \partial y^2} \right) \psi (x,y) + k^2 \psi (x,y) = V \psi (x,y) \,}$ where ${\color {Black} V }$ is a constant using conditions on the behaviour of ${\color {Black} \psi (x,y) . }$

2. You will be able to separate the wave function into products of functions dependent on only one variable and write two one dimensional Schrödinger equations connected to each other through the total energy.

3. You will be able to calculate the transition frequencies for a particle in a two or three dimensional box if the quantum state of the particle is changed by external influence from one allowed state to another allowed state.

4. You will be prompted to read more enlightening articles associated with the concepts presented here from scientific journals.

4. By solving the quizzes, the assignments by yourself and by working out detailed problems for which solutions are given here, on your own, you will learn how to solve and extend a simple model problem to solving the Schrödinger equation for other more complex problems.

Lecture Summary

1. The simplest model problem for solving and following the consequences of the Schrödinger equation, namely the potential-free particle in a two and three dimensional finite regions with infinite barrier at the boundaries is described.

2. Solutions are obtained using the method of separation of variables and provided in simple functional form and are graphically illustrated.

3. The solution is interpreted in terms of the square of the wave functions being associated with the probability density of locating the particle in a given space.

### The Model

It is similar to the one dimensional model in the previous lecture except that the particle can move in a (two- dimensional) plane or a three dimensional volume. Initially we consider two variables and a three dimensional representation for plotting the functions. Therefore the particle is assumed to have only kinetic energy inside the box and is forbidden from escaping from the box due to infinite repulsive potential energies on all sides of the wall. Mathematically the potential is expressed as

${\color {Black} V(x,y) = 0{\rm{ \, \, for }}\, \, 0 < \,x \,< L,{\rm{ }}\, \, 0 < y < L}$

${\color {Black} V(x,y) = \infty {\rm{ \, \, otherwise}} }$

For illustration, we use a square planar region for confining the particle motion in two dimensions. It is not necessary and can be rectangular as well

${\color {Black} (0 < x < L_1 ,{\rm{ }}\, \, 0 < y < L_2 ,{\rm{ \, \, for \, \, example)}} }$

The wave function and the Schrödinger equation have two coordinates as variables (x and y, mutually orthogonal). The two dimensional Schrödinger equation for the particle in this new box is given by

${\color {Black}{ - \frac{{\hbar ^2 }}{{2m}}\left[ {\frac{{\partial ^2 }}{{\partial x^2 }} + \frac{{\partial ^2 }}{{\partial y^2 }}} \right]\Psi (x,y) = \, \, E\Psi (x,y)} }$

(The box appears three dimensional while the motion is in a two dimensional plane. The third dimension is useful for plotting functions of two variables) This equation is obtained in exactly the same manner as was done for one dimensional system. The momentum of the particle in two dimensions has two components, namely x and y components in the xy coordinate system. Therefore, there are two quantum mechanical operators corresponding to these. They result is the above equation. Convince yourself that

${\color {Black} {\frac{{p^2 }}{{2m}} = \frac{{\vec p.\vec p}}{{2m}} = \frac{1}{{2m}}\left( {p_x ^2 + p_y ^2 } \right)} }$

is replaced by the corresponding quantum mechanical operators

${\color {Black}{p_x \Rightarrow - i\hbar \frac{\partial }{{\partial x}},{\rm{ }}\, \, p_y \Rightarrow - i\hbar \frac{\partial }{{\partial y}}} }$

to give the Schrödinger equation.

### Solution of the two dimensional partial differential equation

Use the method of separation of variables. Now for the trick: Propose the wave function

${\color {Black}{\Psi (x,y)} }$

to be a product of two functions, a function of x alone and a function of y alone, i.e.,

${\color {Black} {\Psi (x,y) = X(x).Y(y)} }$

Substitute:

${\color {Black} { - \frac{{\hbar ^2 }}{{2m}}\left[ {\frac{{\partial ^2 }}{{\partial x^2 }} + \frac{{\partial ^2 }}{{\partial y^2 }}} \right]{\rm{ }}X(x)Y(y) = E{\rm{ }}X(x)Y(y)} }$

What is the result?

${\color {Black} { - \frac{{\hbar ^2 }}{{2m}}\left[ {Y(y)\frac{{d^2 X(x)}}{{dx^2 }} + X(x)\frac{{d^2 Y(y)}}{{dy^2 }}} \right] =\, \, E{\rm{ }}X(x)Y(y)} }$

This is because the partial derivative with respect to x will affect only the function dependent on x and the partial derivative with respect to y will affect only the function dependent on y. They are both total derivatives in their respective coordinates. Divide by ${\color {Black}X(x)Y(y) }$ on both sides and get the following:

${\color {Black} { - \frac{{\hbar ^2 }}{{2m}}\left[ {\frac{1}{{X(x)}}\frac{{d^2 X(x)}}{{dx^2 }} + \frac{1}{{Y(y)}}\frac{{d^2 Y(y)}}{{dy^2 }}} \right] =\, E} }$

The first term depends only on x and takes as many values as there are x. The second term depends only on y; and takes as many values as there are y. Each term can be varied independently of the other. Therefore, the above equation is satisfied only when each of the term is independent of the other and is separately equal to a constant.

${\color {Black} { - \frac{1}{{X(x)}}\frac{{\hbar ^2 }}{{2m}}\frac{{d^2 X(x)}}{{dx^2 }} = E_1 {\rm{, \, \, or, \, \, }} - \frac{{\hbar ^2 }}{{2m}}\frac{{d^2 X(x)}}{{dx^2 }} = \, E_1 X(x)} }$

and

${\color {Black} { - \frac{1}{{Y(y)}}\frac{{\hbar ^2 }}{{2m}}\frac{{d^2 Y(y)}}{{dy^2 }} = E_2 {\rm{, \, \, or, \, \, }} - \frac{{\hbar ^2 }}{{2m}}\frac{{d^2 Y(y)}}{{dy^2 }} = \,E_2 Y(y)} }$

${\color {Black} E_1 }$ and ${\color {Black}E_2 }$ satisfy the condition ${\color {Black} {E_1 + E_2 = E} }$ .

What are the new results for the particle in the two dimensional (coordinate) box? Take a square box whose side is of length L. Then we have all the solutions same as that for the one dimensional box except that they are repeated for one coordinate each in the two dimensional problem. An important new result in addition is the requirement that the energies have to add up to a total energy, which will lead to the idea of degeneracy of states, to be discussed below.

${\color {Black} {X_{\color {Black}{n_1 }} (x) = \sqrt {\frac{2}{L}} {\rm{ \, \, sin}}\left ( {\frac {{\color {Black}n_1} {\pi x}}{L}} \right),{\rm{ }}\, \, E_{\color {Black}{n_1 }} \, \, = \frac{{h^2 {\color {Black}n_1 ^2 }}}{{8mL^2 }}} }$

${\color {Black}{Y_{\color {Black}{n_2 }} \, (y) = \sqrt {\frac{2}{L}} {\rm{\, \, sin}}\left( {\frac{{ {\color {Black}n_2} \pi y}}{L}} \right),{\rm{ }}\, \, E_{\color {Black}{n_2 }}\, = \frac{{h^2 {\color {Black}n_2 ^2} }}{{8mL^2 }}} }$

${\color {Black} {E_{\color {Black}{n_1 }} \, \, + \, \, E_{\color {Black}{n_2 }} = \frac{{h^2 }}{{8mL^2 }}\left( {\color {Black}{n_1 ^2 + n_2 ^2 }} \right) = \, \, E_{\color {Black}{n_1 n_2 }} } }$

${\color {Black}{\Psi _{\color {Black}{n_1 n_2 }}(x,y) = X_{\color {Black}{n_1 }} (x) Y_ {\color {Black}{n_2 }} \,\, (y) } }$

Examples of quantum numbers, wave functions and energies

 ${\color {Black} {n_1 = 1,n_2 = 1} }$ ${\color {Black}{\psi _{\color {Black}{11}} (x,y) = \frac{2}{L}\, \, \sin \left( {\frac{{\pi x}}{L}} \right)\, \, \sin \left( {\frac{{\pi y}}{L}} \right)} }$ ${\color {Black}{E_{\color {Black}{11}} = \frac{{h^2 }}{{8mL^2 }}{\color {Black}(1^2 + 1^2 )} = \frac{{h^2 }}{{4mL^2 }}} }$ ${\color {Black} {n_1 \, = 2,n_2\, = 1} }$ ${\color {Black} {\psi _{\color {Black}{\color {Black}{21}}} (x,y) = \frac{2}{L}\, \, \sin \left( {\frac{{{\color {Black}2 } \pi x}}{L}} \right)\, \, \sin \left( {\frac{{\pi y}}{L}} \right)} }$ ${\color {Black}{E_{\color {Black}{21}} = \frac{{h^2 }}{{8mL^2 }}{\color {Black}(2^2 + 1^2 )} = \frac{{5h^2 }}{{8mL^2 }}} }$ ${\color {Black} {n_1 = 1,n_2 = 2} }$ ${\color {Black} {\psi _{\color {Black}{12}} (x,y) = \frac{2}{L}\, \, \sin \left( {\frac{{\pi x}}{L}} \right)\, \, \sin \left( {\frac{{{\color {Black}2}\pi y}}{L}} \right)} }$ ${\color {Black} {E_{\color {Black}{12}} = \frac{{h^2 }}{{8mL^2 }}{\color {Black} (1^2 + 2^2 ) }= \frac{{5h^2 }}{{8mL^2 }}} }$ ${\color {Black} {n_1 = 2,n_2 = 2} }$ ${\color {Black} {\psi _{\color {Black}{22}} (x,y) = \frac{\color {Black}{2}}{L}\, \, \sin \left( {\frac{{{\color {Black}2}\pi x}}{L}} \right)\, \, \sin \left( {\frac{{2\pi y}}{L}} \right)} }$ ${\color {Black} {E_{\color {Black}{22}} = \frac{{h^2 }}{{8mL^2 }}{\color {Black}(2^2 + 2^2 )} = \frac{{h^2 }}{{mL^2 }}} }$ ${\color {Black} {n_1 = 3,n_2 = 2} }$ ${\color {Black} {\psi _{\color {Black}{32}} (x,y) = \frac{2}{L}\, \, \sin \left( {\frac{{{\color {Black}3}\pi x}}{L}} \right)\, \, \sin \left( {\frac{{{\color {Black}2}\pi y}}{L}} \right)} }$ ${\color {Black} {E_{\color {Black}{32}} = \frac{{h^2 }}{{8mL^2 }}{\color {Black}(3^2 + 2^2 )} = \frac{{13h^2 }}{{8mL^2 }}} }$ ${\color {Black} {n_1 = 2,n_2 = 3} }$ ${\color {Black}{\psi _{\color {Black}{23}} (x,y) = \frac{2}{L}\sin \left( {\frac{{{\color {Black}2}\pi x}}{L}} \right)\sin \left( {\frac{{{\color {Black}3}\pi y}}{L}} \right)} }$ ${\color {Black} {E_{\color {Black}{23}} = \frac{{h^2 }}{{8mL^2 }}{\color {Black} (2^2 + 3^2 ) }= \frac{{13h^2 }}{{8mL^2 }}} }$

Thus, i) the particle not having any potential energy inside a square region, ii) moving in a two dimensional plane, iii) confined to move in a square box whose length is L and whose all sides have infinite repulsive potential and iv) obeying the Schrodinger equation has its energy quantized in terms of two quantum numbers ${\color {Black} {n_1}}$ and ${\color {Black}{n_2}}$ such that they determine the total energy of the particle which is now discrete and can have multiple states having the same energy.

What is new besides an additional quantum number?

ANSWER: DEGENERACY (Same energy for more than one possible state of the particle)

${\color {Black}{\Psi _{\color {Black}{n_1 n_2 }} (x,y) = X_{\color {Black}{n_1 }} (x)\, \, Y_{\color {Black}{ n_2 } }(y)}}$

${\color {Black}{\Psi _{\color {Black}{n_2 n_1 }} (x,y) = X_{\color {Black}{n_2 }} (x)\, \, Y_ {\color {Black}{n_1 }} (y)} }$

Two questions for you. Are they two different states of the particle? What about their energy? Degenerate energy levels in units of ${\color {Black} \frac{{h^2 }}{{8mL^2 }} }$ of a particle in a two-d box energy levels and three dimensional graphical representation of a few wave functions and their absolute squares are included in the lecture.

### Plots of wave functions and squares of wave functions

The function plot appears as follows:

 ${\color {Black}{\psi _{\color {Black}{11}} (x,y) = \frac{2}{L}\, \, \sin \left( {\frac{{\pi x}}{L}} \right)\, \, \sin \left( {\frac{{\pi y}}{L}} \right)} }$ ${\color {Black}{\psi^2 _{\color {Black}{11}} (x,y) = \frac{4}{{L^2 }}\sin ^2 \left( {\frac{{\pi x}}{L}} \right)\sin ^2 \left( {\frac{{\pi y}}{L}} \right)} }$ Projection on the x-axis Projection on the x-axis
 Projection on the y-axis Projection on the y-axis ${\color {Black} {\psi _{\color {Black}{12}} (x,y) = \frac{2}{L}\sin \left( {\frac{{\pi x}}{L}} \right)\sin \left( {\frac{{2\pi y}}{L}} \right)} }$ ${\color {Black}{\psi^2 _{\color {Black}{12}} (x,y) = \frac{4}{{L^2 }}\sin ^2 \left( {\frac{{\pi x}}{L}} \right)\sin ^2 \left( {\frac{{2\pi y}}{L}} \right)} }$
 ${\color {Black} {\psi _{\color {Black}{21}} (x,y) = \frac{2}{L}\sin \left( {\frac{{2\pi x}}{L}} \right)\sin \left( {\frac{{\pi y}}{L}} \right)} }$ ${\color {Black}{\psi^2 _{\color {Black}{21}} (x,y) = \frac{4}{{L^2 }}\sin ^2 \left( {\frac{{2\pi x}}{L}} \right)\sin ^2 \left( {\frac{{\pi y}}{L}} \right)} }$ ${\color {Black} {\psi _{\color {Black}{13}} (x,y) = \frac{2}{L}\sin \left( {\frac{{\pi x}}{L}} \right)\sin \left( {\frac{{3\pi y}}{L}} \right)} }$ ${\color {Black}{\psi^2 _{\color {Black}{13}} (x,y) = \frac{4}{{L^2 }}\sin ^2 \left( {\frac{{\pi x}}{L}} \right)\sin ^2 \left( {\frac{{3\pi y}}{L}} \right)} }$
 ${\color {Black} {\psi _{\color {Black}{31}} (x,y) = \frac{2}{L}\sin \left( {\frac{{3\pi x}}{L}} \right)\sin \left( {\frac{{\pi y}}{L}} \right)} }$ ${\color {Black} {\psi^2 _{\color {Black}{31}} (x,y) = \frac{4}{{L^2 }}\sin ^2 \left( {\frac{{3\pi x}}{L}} \right)\sin ^2 \left( {\frac{{\pi y}}{L}} \right)} }$ ${\color {Black} {\psi _{\color {Black}{22}} (x,y) = \frac{2}{L}\sin \left( {\frac{{2\pi x}}{L}} \right)\sin \left( {\frac{{2\pi y}}{L}} \right)} }$ ${\color {Black}{\psi^2 _{\color {Black}{22}} (x,y) = \frac{4}{{L^2 }}\sin ^2 \left( {\frac{{2\pi x}}{L}} \right)\sin ^2 \left( {\frac{{2\pi y}}{L}} \right)} }$
 ${\color {Black} {\psi _{\color {Black}{23}} (x,y) = \frac{2}{L}\sin \left( {\frac{{2\pi x}}{L}} \right)\sin \left( {\frac{{3\pi y}}{L}} \right)} }$ ${\color {Black} {\psi^2 _{\color {Black}{23}} (x,y) = \frac{4}{{L^2 }}\sin ^2 \left( {\frac{{2\pi x}}{L}} \right)\sin ^2 \left( {\frac{{3\pi y}}{L}} \right)} }$ ${\color {Black} {\psi _{\color {Black}{32}} (x,y) = \frac{2}{L}\sin \left( {\frac{{3\pi x}}{L}} \right)\sin \left( {\frac{{2\pi y}}{L}} \right)} }$ ${\color {Black} {\psi^2 _{\color {Black}{32}} (x,y) = \frac{4}{{L^2 }}\sin ^2 \left( {\frac{{3\pi x}}{L}} \right)\sin ^2 \left( {\frac{{2\pi y}}{L}} \right)} }$
 ${\color {Black} {\psi _{\color {Black}{33}} (x,y) = \frac{2}{L}\sin \left( {\frac{{3\pi x}}{L}} \right)\sin \left( {\frac{{3\pi y}}{L}} \right)} }$ ${\color {Black} {\psi^2 _{\color {Black}{33}} (x,y) = \frac{4}{{L^2 }}\sin ^2 \left( {\frac{{3\pi x}}{L}} \right)\sin ^2 \left( {\frac{{3\pi y}}{L}} \right)} }$

### Extension to three dimensions

The extension to describing the particle position in a three dimensional region is now obvious. In cartesian coordinates, the potential is expressed as

${\color {Black} V(x,y,z) = 0{\rm{ \, \, for }}\, \, 0 < \,x \,< L,{\rm{ }}\, \, 0 < y < L,{\rm{ }}\, \, 0 < z < L}$

${\color {Black} V(x,y,z) = \infty {\rm{ \, \, otherwise}} }$

or, using different lengths in each direction, the region can be rectangular and not cubic.

${\color {Black} (0 < x < L_1 ,{\rm{ }}\, \, 0 < y < L_2 , {\rm{ }}\, \, 0 < z < L_3 {\rm{ \, \, for \, \, example)}} }$

The wave function and the Schrödinger equation have three coordinates as variables ($x, \, y \, {\rm{and}} \,z$) which are mutually orthogonal). The three dimensional Schrödinger equation for the particle in this new box is given by

${\color {Black}{ - \frac{\hbar ^2 }{2m}}\left[ \frac{\partial ^2 }{\partial x^2 } + \frac{\partial ^2 }{\partial y^2 } + \frac{\partial ^2 }{\partial z^2 } \right]\Psi (x,y,z) = \, \, E\Psi (x,y,z) }$

Use the method of separation of variables. Propose the wave function to be a product of three functions, a function of x alone, a function of y alone and a function of z alone i.e.,

${\color {Black} {\Psi (x,y,z) = X(x).Y(y).Z(z)} }$

${\color {Black}{\Psi (x,y,z)} }$. This will lead to the separation of the equation into three equations, namely,

${\color {Black} { - \frac{1}{{X(x)}}\frac{{\hbar ^2 }}{{2m}}\frac{{d^2 X(x)}}{{dx^2 }} = E_1 {\rm{, \, \, or, \, \, }} - \frac{{\hbar ^2 }}{{2m}}\frac{{d^2 X(x)}}{{dx^2 }} = \, E_1 X(x)} }$ ,

${\color {Black} { - \frac{1}{{Y(y)}}\frac{{\hbar ^2 }}{{2m}}\frac{{d^2 Y(y)}}{{dy^2 }} = E_2 {\rm{, \, \, or, \, \, }} - \frac{{\hbar ^2 }}{{2m}}\frac{{d^2 Y(y)}}{{dy^2 }} = \,E_2 Y(y)} }$

${\color {Black} { - \frac{1}{{Z(z)}}\frac{{\hbar ^2 }}{{2m}}\frac{{d^2 Z(z)}}{{dz^2 }} = E_3 {\rm{, \, \, or, \, \, }} - \frac{{\hbar ^2 }}{{2m}}\frac{{d^2 Z(z)}}{{dz^2 }} = \,E_3 Z(z)} }$ and

${\color {Black} {E_1 + E_2 + E_3 = E} }$ .

The solutions are given in exactly the same way with one more for the z coordinate in the third dimension,

${\color {Black} {X_{\color {Black}{n_1 }} (x) = \sqrt {\frac{2}{L}} {\rm{ \, \, sin}}\left ( {\frac {{\color {Black}n_1} {\pi x}}{L}} \right),{\rm{ }}\, \, E_{\color {Black}{n_1 }} \, \, = \frac{{h^2 {\color {Black}n_1 ^2 }}}{{8mL^2 }}} }$

${\color {Black}{Y_{\color {Black}{n_2 }} \, (y) = \sqrt {\frac{2}{L}} {\rm{\, \, sin}}\left( {\frac{{ {\color {Black}n_2} \pi y}}{L}} \right),{\rm{ }}\, \, E_{\color {Black}{n_2 }}\, = \frac{{h^2 {\color {Black}n_2 ^2} }}{{8mL^2 }}} }$

${\color {Black}{Z_{\color {Black}{n_3 }} \, (z) = \sqrt {\frac{2}{L}} {\rm{\, \, sin}}\left( {\frac{{ {\color {Black}n_3} \pi z}}{L}} \right),{\rm{ }}\, \, E_{\color {Black}{n_3 }}\, = \frac{{h^2 {\color {Black}n_3 ^2} }}{{8mL^2 }}} }$

$E_{n_1} \, \, + \, \, E_{n_2} + \, \, E_{n_3} = \frac{h^2}{8mL^2} \left( n_1 ^2 + n_2 ^2 +n_3 ^2 \right) = \, \, E_{n_1 n_2 n_3}$

$\Psi _{n_1 n_2 n_3}(x,y,z) = X_{n_1} (x) Y_{n_2} (y) Z_{n_3 }(z)$

### Postulates of quantum mechanics

1. The state of any physical system is described in quantum mechanics by one or more wave functions. The wave function is a function of time and coordinates of all parts comprising the system. It is in general, a complex function denoted by ${\color {Black} \psi }$ and has the following interpretation. The product ${\color {Black} \psi *\psi d\tau }$ where ${\color {Black}d\tau }$ is an infinitesimally small volume available to the system, is the probability of finding the system in that volume.

2. To every physically measurable quantity in classical mechanics there corresponds a linear operator in quantum mechanics. For all measurable quantities, the operator is hermitian. Operators act on wave functions to give either the wave functions back or other functions. Examples of linear operators in quantum mechanics are position, momentum, angular momentum, kinetic energy, total energy etc. One of the most important operators is the momentum ${\color {Black}\hat{p}_x }$ of a particle whose position coordinate is given by x

${\color {Black} {\hat{p}_x = - i\hbar \frac{\partial}{{\partial x}}} }$

In general if the coordinate is given by q , the momentum operator ${\color {Black} \hat{p}_q }$ is given by

${\color {Black} {\hat{p}_q = - i\hbar \frac{\partial }{{\partial q}}} }$

a partial derivative with respect to one coordinate, keeping all other variables fixed.

3. The only observable (physically measurable) quantities are the quantities called the eigenvalues of the operators. The eigenvalue equation for an operator is given as:

${\color {Black}{\hat{A} \phi = a\phi } }$

Where ${\color {Black} \hat{A} }$ is the operator acting on a function ${\color {Black} \phi }$ which results in the function being multiplied by a constant a, called, its eigenvalue. The Schrödinger equation you solve is an eigenvalue equation for the total energy operator known as the Hamiltonian.

4. The wave functions satisfy the Schrödinger equation which is to be taken as the fundamental equation of motion in quantum mechanics. The time dependent Schrödinger equation is given as ${\color {Black} {i\hbar \frac{{\partial \Psi }}{{\partial t}} = \hat{H} \Psi } }$

where $\hat{H}$ is the Hamiltonian operator, the operator for the total energy. The time- independent Schrodinger equation is given by ${\color {Black} {\hat{H} \Psi = E\Psi } }$

which is an eigenvalue equation for the total energy operator.

5. This postulate is concerned with the expectation values of operators of systems in a state ψ . When the system is in a state represented by a wave function ψ, the average value (the expected or expectation value) of any operator associated with a physically measurable quantity is given by the formula ${\color {Black} {\left\langle {\hat A} \right\rangle = \frac{{\int {\Psi ^*} \hat{} A\Psi d\tau }}{{\int {\Psi^ *} \Psi d\tau }}} }$

The integration must be carried out over all the space available to the system.